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Neko [114]
3 years ago
8

An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

rrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull's-eye if its initial speed is 37 m/s?
Physics
1 answer:
Liono4ka [1.6K]3 years ago
4 0

Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

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Answer:

nerve pathways

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3 0
3 years ago
Please Help!!!!
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Answer:

v = 0.92 c

Explanation:

Here, we will use the time dilation formula from Einstein's theory of relativity to find the speed of traveling of the friend:

t =\frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}} \\\\\\\sqrt{1-\frac{v^2}{c^2}}=\frac{t_o}{t}\\\\

where,

v = speed of traveling = ?

c = speed of light

t = time of return = 10 years

t₀ = time passed on earth = 4 years

Therefore,

\sqrt{1-\frac{v^2}{c^2}} = \frac{4\ years}{10\ years}\\\\  1-\frac{v^2}{c^2}=(\frac{2}{5})^2\\\\\frac{v^2}{c^2} = 1-\frac{4}{25}\\\\\frac{v^2}{c^2} = \frac{21}{25}\\\\v^2 = 0.84c^2\\\\

<u>v = 0.92 c</u>

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When the pump extracts the air from the bell jar, the pressure inside the balloon naturally decreases. The balloon usually has a air pressure around it, which restricts its size, but when this air is extracted and the pressure around it decreases the gas in the balloon will expand and the balloon seems to be inflating. When you release the air back into the bell jar, it will once again compress back to its actual size.

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